A spring is hung from the ceiling. A 0.300 kg block is then attached to the free end of the spring. When released from rest, the block drops 0.170 m before momentarily coming to rest.

(a) What is the spring constant of the spring?

(b) Find the angular frequency of the block's vibrations.

a) F=-kx => k=F/x => k=mg/.5x

.300kg(9.8)/.5(.170m)= k

b) w=squareroot(k/m)

To find the spring constant of the spring, you can use Hooke's Law, which states that the force exerted by a spring is directly proportional to the displacement of the spring from its equilibrium position.

(a) To find the spring constant, you can use the formula:

F = kx

where F is the force exerted by the spring, k is the spring constant, and x is the displacement from the equilibrium position.

In the given problem, the block drops 0.170 m before momentarily coming to rest. At this point, the block is at its maximum displacement from the equilibrium position. At this point, the force exerted by the spring balances the force of gravity acting on the block.

The force of gravity acting on the block can be calculated using Newton's second law:

Fg = mg

where Fg is the force of gravity, m is the mass of the block, and g is the acceleration due to gravity.

In this case, the mass of the block is 0.300 kg, and the acceleration due to gravity is approximately 9.8 m/s^2. Therefore,

Fg = (0.300 kg)(9.8 m/s^2) = 2.94 N

Since the force exerted by the spring balances the force of gravity, we have:

F = 2.94 N

The displacement of the spring from the equilibrium position is 0.170 m. Therefore,

kx = 2.94 N

k(0.170 m) = 2.94 N

Solving for k, we have:

k = 2.94 N / (0.170 m) ≈ 17.29 N/m

So, the spring constant of the spring is approximately 17.29 N/m.

(b) The angular frequency of the block's vibrations can be calculated using the formula:

ω = √(k/m)

where ω is the angular frequency, k is the spring constant, and m is the mass of the block.

In this case, the spring constant is 17.29 N/m, and the mass of the block is 0.300 kg. Therefore,

ω = √(17.29 N/m / 0.300 kg)

ω = √(57.63 N/m^2/kg)

ω ≈ 7.6 rad/s

So, the angular frequency of the block's vibrations is approximately 7.6 rad/s.